A090365 - OEIS

%I #93 Apr 26 2024 06:59:01

%S 1,1,3,11,47,225,1177,6625,39723,251939,1681535,11764185,86002177,

%T 655305697,5193232611,42726002123,364338045647,3215471252769,

%U 29331858429241,276224445794785,2682395337435723,26832698102762435,276221586866499839,2923468922184615897

%N Shifts 1 place left under the INVERT transform of the BINOMIAL transform of this sequence.

%C The Hankel transform of this sequence is A000178(n+1); example: det([1,1,3; 1,3,11; 3,11,47]) = 12. - _Philippe Deléham_, Mar 02 2005

%C a(n) appears to be the number of indecomposable permutations (A003319) of [n+1] that avoid both of the dashed patterns 32-41 and 41-32. - _David Callan_, Aug 27 2014

%C This is true: A nonempty permutation avoids 32-41 and 41-32 if and only if all its components do so. So if A(x) denotes the g.f. for indecomposable {32-41,41-32}-avoiders, then F(x):=1/(1-A(x)) is the g.f. for all {32-41,41-32}-avoiders. From A074664, F(x)=1/x(1-1/B(x)) where B(x) is the o.g.f. for the Bell numbers. Solve for A(x). - _David Callan_, Jul 21 2017

%H Alois P. Heinz, <a href="/A090365/b090365.txt">Table of n, a(n) for n = 0..574</a>

%F G.f.: A(x) satisfies A(x) = 1/(1 - A(x/(1-x))*x/(1-x) ).

%F a(n) = Sum_{k = 0..n} A085838(n, k). - _Philippe Deléham_, Jun 04 2004

%F G.f.: 1/x-1-1/(B(x)-1) where B(x) = g.f. for A000110 the Bell numbers. - _Vladeta Jovovic_, Aug 08 2004

%F a(n) = Sum_{k=0..n} A094456(n,k). - _Philippe Deléham_, Nov 07 2007

%F G.f.: 1/(1-x/(1-2x/(1-x/(1-3x/(1-x/(1-4x/(1-x/(1-5x/(1-... (continued fraction). - _Paul Barry_, Feb 25 2010

%F From _Sergei N. Gladkovskii_, Jan 06 2012  - May 12 2013: (Start)

%F Continued fractions:

%F G.f.: 1 - x/(G(0)+x); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1).

%F G.f.: 1/x - 1/2 + (x^2-4)/(4*U(0)-2*x^2+8) where U(k) = k*(2*k+3)*x^2 + x - 2 - (2-x+2*k*x)*(2+3*x+2*k*x)*(k+1)*x^2/U(k+1).

%F G.f.: 1/x+1/(U(0)-1) where U(k) = -x*k + 1 - x - x^2*(k+1)/U(k+1).

%F G.f.: (1 - U(0))/x - 1 where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1).

%F G.f.: (1 - U(0))/x where U(k) = 1 - x*(k+1)/(1-x/U(k+1)).

%F G.f.: 1/x + 1/( G(0)-1) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/ G(k+1) ))).

%F G.f.:1/x + 1/( G(0) - 1 ) where G(k) = 1 - x/(1 - x*(k+1)/G(k+1) ).

%F G.f.: (1 - Q(0))/x where Q(k) = 1 + x/(x*k - 1 )/Q(k+1).

%F G.f.: 1/x - 1/x/Q(0), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))).

%F (End)

%F Conjecture: a(n) = b(2^(n-1) - 1) for n > 0 with a(0) = 1 where b(n) = b((n - 2^f(n))/2) + b(floor((2n - 2^f(n))/2)) + b(A025480(n-1)) for n > 0 with b(0) = 1 and where f(n) = A007814(n). - _Mikhail Kurkov_, Jan 11 2022

%p bintr:= proc(p) proc(n) add(p(k) *binomial(n,k), k=0..n) end end:

%p invtr:= proc(p) local b;

%p            b:= proc(n) option remember; local i;

%p                 `if`(n<1, 1, add(b(n-i) *p(i-1), i=1..n+1))

%p                end;

%p         end:

%p b:= invtr(bintr(a)):

%p a:= n-> `if`(n<0, 0, b(n-1)):

%p seq(a(n), n=0..30);  # _Alois P. Heinz_, Jun 28 2012

%t a[n_] := Module[{A, B}, A = 1+x; For[k=1, k <= n, k++, B = (A /. x -> x/(1 - x))/(1-x) + O[x]^n // Normal; A = 1 + x*A*B]; SeriesCoefficient[A, {x, 0, n}]]; Table[a[n], {n, 0, 23}] (* _Jean-François Alcover_, Oct 23 2016, adapted from PARI *)

%o (PARI) {a(n)=local(A); if(n<0,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x, x/(1-x))/(1-x)+x*O(x^n); A=1+x*A*B);polcoeff(A,n,x))}

%Y Cf. A090366, A090367.

%Y Cf. A074664.

%Y Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A329369, A341392, A347204, A347205.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Nov 26 2003